How Do You Correctly Solve the Limit of (6π−6x)tan(x/2) as x Approaches π?

[JackieAnne]

Find the limit. limx->pi- (6pi−6x)tan(x/2) =

lim(x→π) [ (6π - 6x) tan(x/2) ]
= 6 lim(x→π) [ (π - x) tan(x/2) ]
= 6 lim(x→π) [ (π - x) / cot(x/2) ]
apply l'Hôpital's Rule since it's in the form of 0/0, at x = π
= 6lim(x→π) [ d/dx (π - x) / d/dx (cot(x/2)) ]
= 6lim(x→π) [ -1 / ((1/2) (-csc²(x/2)) ]
= 6lim(x→π) [ 2 / csc²(x/2) ]
= 6lim(x→π) [ 2 sin²(x/2) ]
= 12lim(x→π) [ sin²(x/2) ]
= 12 [ sin²(π/2) ]
= 12 [ 1/2 ]
= 1/6

This answer is incorrect. Can anyone indicate where I went wrong?

 

[JonF]

sin²(π/2) = 1, and 12(1/2) = 6.

Also this would have been much easier if you wrote tan(x/2) as sin(x/2)/cos(x/2)